Lewis Structures and Molecular Shapes of NF3, PF5, CO2-, CO2, CO, CF4, XeF4, and ClF3, Study Guides, Projects, Research of Geometry

Lewis electron-dot structures and explanations of the bonding and molecular shapes for various fluorine and nitrogen compounds, including nf3, pf5, co2-, co2, co, cf4, xef4, and clf3. It covers topics such as resonance structures, bond lengths, and hybridization.

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2021/2022

Uploaded on 09/12/2022

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Bonding Set I
1. NF3 and PF5 are stable molecules. Write the Lewis electron-dot formulas for these molecules. On the
basis of structural and bonding considerations, account for the fact that NF3 and PF5 are stable molecules
but NF5 does not exist.
2. (a) Draw the Lewis electron-dot structures for CO32-, CO2, and CO, including resonance structures
where appropriate.
(b) Which of the three species has the shortest C-O bond length? Explain the reason for your answer.
(c) Predict the molecular shapes for the three species. Explain how you arrived at your predictions.
3. CF4 XeF4 ClF3
(a) Draw a Lewis electron-dot structure for each of the molecules above and identify the shape of each.
(b) Use the valence shell electron-pair repulsion (VSEPR) model to explain the geometry of each of
these molecules.
4. Answer the following questions about the structures of ions that contain only sulfur and fluorine.
(a) The compounds SF4 and BF3 react to form an ionic compound according to the following
equation.
SF4 + BF3 SF3BF4
(i) Draw a complete Lewis structure for the SF3+ cation in SF3BF4.
(ii) Identify the type of hybridization exhibited by sulfur in the SF3+ cation.
(iii) Identify the geometry of the SF3+ cation that is consistent with the Lewis structure drawn in
part (a)(i).
(iv) Predict whether the F—S—F bond angle in the SF3+ cation is larger than, equal to, or
smaller than 109.50˚. Justify your answer.
(b) The compounds SF4 and CsF react to form an ionic compound according to the following
equation.
SF4 + CsF CsSF5
(i) Draw a complete Lewis structure for the SF5 anion in CsSF5.
(ii) Identify the type of hybridization exhibited by sulfur in the SF5 anion.
(iii) Identify the geometry of the SF5 anion that is consistent with the Lewis structure drawn in
part (b)(i).
(iv) Identify the oxidation number of sulfur in the compound CsSF5.
pf3

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Bonding Set I

  1. NF 3 and PF 5 are stable molecules. Write the Lewis electron-dot formulas for these molecules. On the basis of structural and bonding considerations, account for the fact that NF 3 and PF 5 are stable molecules but NF 5 does not exist.
  2. (a) Draw the Lewis electron-dot structures for CO 3 2-, CO 2 , and CO, including resonance structures where appropriate.

(b) Which of the three species has the shortest C-O bond length? Explain the reason for your answer. (c) Predict the molecular shapes for the three species. Explain how you arrived at your predictions.

  1. CF 4 XeF 4 ClF (^3)

(a) Draw a Lewis electron-dot structure for each of the molecules above and identify the shape of each.

(b) Use the valence shell electron-pair repulsion (VSEPR) model to explain the geometry of each of these molecules.

  1. Answer the following questions about the structures of ions that contain only sulfur and fluorine.

(a) The compounds SF 4 and BF 3 react to form an ionic compound according to the following equation. SF 4 + BF 3 → SF 3 BF (^4) (i) Draw a complete Lewis structure for the SF 3 +^ cation in SF 3 BF 4. (ii) Identify the type of hybridization exhibited by sulfur in the SF 3 +^ cation. (iii) Identify the geometry of the SF 3 +^ cation that is consistent with the Lewis structure drawn in part (a)(i). (iv) Predict whether the F—S—F bond angle in the SF 3 +^ cation is larger than, equal to, or smaller than 109.50˚. Justify your answer.

(b) The compounds SF 4 and CsF react to form an ionic compound according to the following equation. SF 4 + CsF → CsSF 5 (i) Draw a complete Lewis structure for the SF 5 –^ anion in CsSF 5. (ii) Identify the type of hybridization exhibited by sulfur in the SF 5 –^ anion. (iii) Identify the geometry of the SF 5 –^ anion that is consistent with the Lewis structure drawn in part (b)(i). (iv) Identify the oxidation number of sulfur in the compound CsSF 5.

Answers

NF 3 PF (^5)

F P

F

F

F F

Your Lewis structures should show the sp 3 hybridization in NF 3 and the sp 3 d hybridization in PF 5. The reason NF 5 doesn’t exist is because N has no d orbitals available for hybridization.

(a)

-2-2 -2-2 (^) -2-

:::: ::::^ :: ::

(b) CO has the shortest bond length because of the triple bond.

(c) CO 3 2-^ is trigonal planar because of the sp^2 hybridization, which puts three sigma bonds in a

plane with lobes of an unhybridized p orbital above and below that plane, forcing the planar

shape. CO 2 has two double bonds, forcing it to be linear (because of the perpendicular pi

bonds). CO has a triple bond, which is also linear for the same reason as CO 2.

  1. (a)

Tetrahedral Square planar T-shaped

(b) CF 4 has 4 shared, no unshared pairs. The maximum angular separation gives a tetrahedron.

XeF 4 has two unshared pairs on opposite sides of the Xe. This forces the remaining F atoms into a central plane.